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Consider a plasma that just's been formed and then left alone. Being far from equilibrium, the plasma will evolve towards an equilibrium state. At this stage, it's not very useful to characterize the plasma with a temperature because the velocity distribution would bear little resemblance to a Boltzmann distribution, or really any kind of distribution ...


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There are lots of different types of plasmas. In a thermal plasma the electrons and ions will have the same temperature. In a non-thermal plasma the discharge is driven by some external power supply e.g. capacitatively coupled RF, inductively coupled, pulsed DC E field etc. In a non-thermal plasma the electrons generally have a higher temperature than ...


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If the string is massless and taut, then the wave velocity is infinite - that is, a component of the force at one mass will immediately be felt at the other mass. But to answer your first question, not all the force of the impact will be transmitted along the string, as made clear by this diagram: As for your second question: momentum for the system is ...


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What acts here is called impulse (of a Force) Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) they collide. B is at rest (v = 0): Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what steel the balls are made you can calculate the ...


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Sure. An elastic collision is merely an idealized process which never occurs in nature. On the microscopic level, the two bodies are made up out of atoms. The electron shells of these atoms contain electrons. Since negative charges repel each other, the two bodys will repel each other in a smooth fashion (an inverse square force is acting).


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One key point is that the velocity and momentum of the center of mass of the system do not change as a result of the collision. I would recommend calculating the velocity of the center of mass prior to the collision, and then use the fact that it is unchanged to compute the new velocities. If the collision is truly elastic (no dissipation), then the kinetic ...


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If you assume linear force displacement relationship (small displacements) with stiffness $k$ then $F(t) = k\, x(t)$ where $x(t)$ is the separation of the two objects. Also initially the impact speed is $V = \dot{x}(0)$. The impact time is characterized by the natural frequency of the system $$\omega = \sqrt{\frac{k}{m}}$$ where $m = \left( \frac{1}{m_1} + ...


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(make comment as an answer) As mentioned in the comment here, the impact forces that are active during the time-frame of the actual impact are 1) unknown 2) difficult to put in analytic form That is why results like the conservation of momentum theorem are used. One can do estimations or approximations of these impact forces but would have to use more ...


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what is the force that the first marble applied one the second marble? The collision is almost instantaneous. Wouldn't that make the force in ΣF=Δmv/Δt insanely large because Δt is so small? Suppose two steel balls A, B of equal mass (m = 0.1 r = 0.03 m) collide and B is at rest: Ball A will exert on b the Impulse of a Force $J$ and its velocity, ...


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As stated in other answers, $\Delta t$, although small than times you're used to perceiving, is not 0. Because the marbles do not deform very much, the collision does not appear to take much time. However, in the grand scheme of things, the change in momentum for each marble during the collision is also small. Although answers vary, a quick google search ...


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The force can be surprisingly large, but $\Delta t$ is not zero, and the force is not infinite. Make some estimates: the duration of the collision is so short that our eyes and brain cannot perceive it. Make an estimate for an upper limit for the duration. (There's no right answer, but a lot of wrong answers. For example, I would think that a duration ...


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In this type of collision where you have what amounts to a very quick change in velocity, the force is called an impulse force and it is best to think of the equation a little differently. For example, instead of: $$ \sum F = \frac{\Delta mv}{\Delta t} $$ Think of $\int F \mathrm{d}t$ being equal to the change in momentum, that is: $$ \Delta mv = \int ...


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The problem with your solution is that the inelastic collision and assumption that kinetic energy is conserved are mutually exclusive. You can see that in your math when you try to solve for $v_2$. Rewriting equation $(1)$ gives $v_1=\left(m+M\right)v_2/m$ which inserted into $(2)$ yields $$ m \left(\frac{m+M}mv_2\right)^2 = \left(m+M\right)v_2^2.$$ This ...


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though momentum as well as energy is conserved but definitely the sum of individual momentum of particles is not equal to sum of individual K.E. of the particles. also there may be different value of K.E. for same momentum. So can not make any result by manipulating the equations.


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Both energy and momentum are conserved as always. But to understand why this statement is true you have to look at the system as you described it a little more closely: In order for blocks A and B to stick together after the collision, the force between them should be zero when the velocity difference is zero -otherwise that force would continue to ...


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Let's make a concrete example with numbers: Suppose that $v_a = 6m/s$ and $v_b = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$ . According to the conservation of energy and momentum: Kinetic energy and momentum are conserved only in a perfect elastic collision, if the bodies stick together the collision is inelastic an ...


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The thing to keep in mind is that for inelastic collisions, energy is not conserved within the system. Some energy is lost as thermal energy, and after all, it takes some energy to get the blocks to stick together. This doesn't mean that the Law of Conservation of Energy is false, though. It just means that energy has left the system that you're studying ...


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IMHO a dust-particle-sized black hole, roughly of the mass of Pluto, moving at few hundreds km/s, could produce effects similar to those are pictured, namely: Ejecta above the entry point (destruction of the crust by the hole’s gravity); Except vicinity of aforementioned point, an almost undamaged leading hemisphere (tidal forces become too weak to break ...


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I propose my solution to this paradox. Well let us assume the insect is travelling at velocity $v$ and train travelling towards the insect at speed $v_1$ (not in insects frame of reference, but trains frame of reference). That said, now the insect hits the train and as a result will decelerate at an finite amount of time $t$ as it is impossible to ...


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That is a very good paradox. But it must have a solution, so let's try and find something that is adequate to our everyday experience: the train must not, obviously, stop. I think the reason why this problem is startling at first is that it refers to an insect. When we think of an insect, we think of something with almost only two dimensions. But an insect, ...


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Both the insect and the train window are deformable. Microscopically so, but deformable nonetheless. Because of that fact, the insect slows continuously to zero, reverses direction, and then speeds up in the direction the train is going. To our human perception this happens imperceptibly fast.


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Yes, the insect does actually travel at 0mph when it reverses direction. The critical aspect that you are missing is that the time that it is not moving is infinitely short. Actually, I would think that the head stops, then the thorax and finally the tail as it is squished up against the train.


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Let's turn this around. In an inelastic collision, some of the energy, instead of remaining with the center of mass of the objects colliding, is dissipated as heat - an increase in the random motion of the atoms and molecules of the objects colliding. At the (sub) atomic level, the two particles involved in a simple collision are the same two objects that ...


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Momentum: p=mv Kinetic energy: E(k) = 1/2mv^2 So the answer is both. Kinetic energy and momentum are both functions of mass and velocity, and so functions of each other. One goes up, the other goes up: from momentum: v = p/m plugging back into Kinetic energy: E(k) = 1/2m[p/m]^2


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Now if they are both shot at ballistic gelatin, which one is expected to cause more damage if both are stopped by the gelatin ? By damage here I mean more penetration, bigger cavity, heat and any other deformation of the gelatin. In other words, if the velocity of a projectile is doubled, will the amount of damage it causes when it collides with ...


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I think I saw a video about this problem... but I couldn't find a link. I believe it has to do with where the bullet hits the "free" object. Whether it hits on a line through the center of mass, or whether it hits near an edge (and thus has both momentum and angular momentum.) Maybe someone else can find the link.


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If you have an elastic collision between objects A and B and where 'kinetic energy is conserved', does this mean object 1 will always have the same velocity it had before the collision? The speed and the direction of A after an elastic collision $v_f$ depends on the mass of B: the ratio $v_f/v_i$ varies from -1 to +1: it is equal to -1 ($v_f = ...


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As gravity force is their in Y direction so momentum is not conserved but in X direction as MC is valid so no change in the velocity. if take the components we can see the impulse is in positive Y direction only i.e. direction of force. so acceleration due to collision is in positive Y direction but gravity is their in negative Y direction. in X direction it ...


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Normally, I think that the surface will react with a force in the OC→ direction Yes, if the ball's force on the surface is in direction A, then by Newton's third law, the surface's force on the wall is in direction C. This is what Newton's third law says. The third law applies to all forces in the same way. However, as pointed out by Jan Hudec in a ...


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The force between two colliding bodies is not in the direction of motion. The normal and parallel components have to be treated separately. The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface ...


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By the way, OA/OB/OC are not force vectors. The only force vector that is acting on the ball before (and after) it hits the ground is its weight, and it is acting vertically downwards. OA/OB/OC are velocity vectors. Try to split OA up into its vertical and horizontal components. When the ball hits the floor, reverse the vertical component due to impulse ...


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Let $v_1$ and $v_2$ be the particle's velocities in the center of mass coordinate system after the collision. by conservation of momentum and energy we have \begin{gather}\tag{1}m_1v_1+m_2v_2=0 \\ \tag{2}m_1v_1^2+m_2v_2^2=m_1(v-v_c)^2+m_2v_c^2 \end{gather} Isolating $v_1$ in $(1)$ and substituting in $(2)$: ...


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$m_2$ will leave with the same magnitude of momentum but opposite direction. Now the assertion is made that in an elastic collision, $m_1$ and $m_2$ have the same speeds leaving the collision as entering it. In other words, the speed of $m_1$ is $v-v_c$ and the speed of $m_2$ is $v_c$ after the collision. In order to simplify things and to ...


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My question, concerning protons, got its answer from this document (Chapter 3), Anomalous $A_{NN}$ Spin-Spin correlation.



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